I know that Γ′(1)=−γ, but how does one prove this?
Starting from the basics, we have that:
Γ(x)=∫∞0e−ttx−1dt
How do we differentiate this? How do we then find that
Γ′(1)=∫∞0e−tlog(t)dt
and how would one solve this integral?
Answer
Taken from this answer:
By the recursive relation Γ(x+1)=xΓ(x), we get
log(Γ(x))=log(Γ(n+x))−log(x)−log(x+1)−log(x+2)−⋯−log(x+n−1)
Differentiating (1) with respect to x, evaluating at x=1, and letting n→∞ yields
Γ′(1)Γ(1)=log(n)+O(1n)−11−12−13−⋯−1n→−γ
A discussion of why ddxlog(Γ(x))=log(x)+O(1x), and a proof of the log-convexity of Γ(x), is given in the answer cited above. It is easy to accept since log(Γ(n+1))=log(Γ(n))+log(n).
Another Approach
∫∞0log(t)e−tdt=limn→∞∫n0log(t)(1−tn)ndt=limn→∞n∫10(log(t)+log(n))(1−t)ndt=limn→∞(nn+1log(n)+n∫10log(1−t)tndt)=limn→∞(nn+1log(n)−nn+1Hn+1)=−γ
Explanation:
(3a): Monotone Convergence and e−t=limn→∞(1−tn)n
(3b): substitute t↦nt
(3c): substitute t↦1−t
(3d): integrate by parts ∫10log(1−t)tndt=−1n+1∫101−tn+11−tdt
(3e): γ=limn→∞Hn−log(n)
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